Chapter 16, Problem 39E

To determine

The required $z$ parameters for the given condition.

Answer to Problem 39E

The required $z$ parameters is $z=\left[\begin{array}{cc}\left(89.71-j98.38\right)& \left(94.06-j4.34\right)\\ \left(527.27+j9401.42\right)& \left(563.86-j35.47\right)\end{array}\right]\text{Ω}$.

Explanation of Solution

Given data:

The angular frequency is $\omega ={10}^8\text{rad}/\text{s}$.

Calculation:

The given diagram is shown in Figure 1.

The conversion from $\text{pF}$ to $\text{F}$ is given by,

$1\text{pF}={10}^{-12}\text{F}$

The conversion from $5\text{pF}$ to $\text{F}$ is given by,

$5\text{pF}=5\times {10}^{-12}5\text{F}$

The conversion from $\text{kΩ}$ to $\text{Ω}$ is given by,

$1\text{kΩ}={10}^3\Omega$

The conversion from $\text{100}\text{kΩ}$ to $\text{Ω}$ is given by,

$\begin{array}{c}100\text{kΩ}=100\times {10}^3\Omega \\ =100000\Omega \end{array}$

The conversion from $\text{10}\text{kΩ}$ to $\text{Ω}$ is given by,

$\begin{array}{c}10\text{kΩ}=10\times {10}^3\Omega \\ =10000\Omega \end{array}$

The capacitive reactance of $5\text{pF}$ capacitor is given by,

$X_{\text{C5}}=\frac{1}{j\omega C_5}$

Substitute ${10}^8\text{rad}/\text{s}$ for $\omega$ and $5\times {10}^{-12}\text{F}$ for $C_5$ in the above equation.

$\begin{array}{c}{X}_{\text{C5}}=\frac{1}{j\left({10}^8\text{rad}/\text{s}\right)\left(5\times {10}^{-12}\text{F}\right)}\\ =-j2000\Omega \end{array}$

The capacitive reactance of $1\text{pF}$ capacitor is given by,

$X_{\text{C1}}=\frac{1}{j\omega C_1}$

Substitute ${10}^8\text{rad}/\text{s}$ for $\omega$ and $1\times {10}^{-12}\text{F}$ for $C_1$ in the above equation.

$\begin{array}{c}{X}_{\text{C1}}=\frac{1}{j\left({10}^8\text{rad}/\text{s}\right)\left(1\times {10}^{-12}\text{F}\right)}\\ =-j10000\Omega \end{array}$

The modified diagram is shown in Figure 2.

Apply KCL at node $V_{\text{A}}$.

$\begin{array}{l}-I_1+\frac{V_{\text{A}}}{100000}+\frac{V_{\text{A}}}{-j2000}+\frac{V_{\text{A}}-V_{\text{B}}}{-j10000}=0\\ I_1={10}^{-5}{V}_{\text{A}}+j\left(5\times {10}^{-4}\right)V_{\text{A}}+j\left(1\times {10}^{-4}\right)\left(V_{\text{A}}-V_{\text{B}}\right)\\ I_1=\left({10}^{-5}+j\left(6\times {10}^{-4}\right)\right)V_{\text{A}}-j\left(1\times {10}^{-4}\right)V_{\text{B}}\end{array}$        (1)

Apply KCL at node $V_{\text{B}}$.

$\begin{array}{l}-I_2+\frac{V_{\text{B}}}{10000}+0.01V_{\text{A}}+\frac{V_{\text{B}}-V_{\text{A}}}{-j10000}=0\\ I_2=\left(1\times {10}^{-4}\right)V_{\text{B}}+0.01V_{\text{A}}+j\left(1\times {10}^{-4}\right)\left(V_{\text{B}}-V_{\text{A}}\right)\\ I_1=\left(0.01-j\left(1\times {10}^{-4}\right)\right)V_{\text{A}}-\left(\left(1\times {10}^{-4}\right)+j\left(1\times {10}^{-4}\right)\right)V_{\text{B}}\end{array}$        (2)

The standard equation for admittance parameters is,

$I_1=y_{11}{V}_{\text{A}}+y_{12}{V}_{\text{A}}$        (3)

$I_2=y_{21}{V}_{\text{A}}+y_{22}{V}_{\text{B}}$        (4)

Write equation (1) and equation (2) in matrix form.

$\left[\begin{array}{c}{I}_1\\ I_2\end{array}\right]=\left[\begin{array}{cc}{10}^{-5}+j\left(6\times {10}^{-4}\right)& -j\left(1\times {10}^{-4}\right)\\ 0.01-j\left(1\times {10}^{-4}\right)& \left(1\times {10}^{-4}\right)+j\left(1\times {10}^{-4}\right)\end{array}\right]\left[\begin{array}{c}{V}_{\text{A}}\\ V_{\text{B}}\end{array}\right]$        (5)

Write equation (3) and equation (4) in matrix form

$\left[\begin{array}{c}{I}_1\\ I_2\end{array}\right]=\left[\begin{array}{cc}{y}_{11}& y_{12}\\ y_{21}& y_{22}\end{array}\right]\left[\begin{array}{c}{V}_{\text{A}}\\ V_{\text{B}}\end{array}\right]$        (6)

Compare equation (5) with equation (6).

$\begin{array}{l}\left[\begin{array}{cc}{y}_{11}& y_{12}\\ y_{21}& y_{22}\end{array}\right]=\left[\begin{array}{cc}{10}^{-5}+j\left(6\times {10}^{-4}\right)& -j\left(1\times {10}^{-4}\right)\\ 0.01-j\left(1\times {10}^{-4}\right)& \left(1\times {10}^{-4}\right)+j\left(1\times {10}^{-4}\right)\end{array}\right]\\ y=\left[\begin{array}{cc}{10}^{-5}+j\left(6\times {10}^{-4}\right)& -j\left(1\times {10}^{-4}\right)\\ 0.01-j\left(1\times {10}^{-4}\right)& \left(1\times {10}^{-4}\right)+j\left(1\times {10}^{-4}\right)\end{array}\right]\end{array}$

The voltage expression $V_1$ by Crammer’s rule is given by,

$V_1=\frac{\left|\begin{array}{cc}{I}_1& y_{12}\\ I_2& y_{22}\end{array}\right|}{\left|\begin{array}{cc}{y}_{11}& y_{12}\\ y_{21}& y_{22}\end{array}\right|}$

Substitute ${10}^{-5}+j\left(6\times {10}^{-4}\right)$ for $y_{11}$, $-j\left(1\times {10}^{-4}\right)$ for $y_{12}$, $0.01-j\left(1\times {10}^{-4}\right)$ for $y_{21}$ and $\left(1\times {10}^{-4}\right)+j\left(1\times {10}^{-4}\right)$ for $y_{22}$ in above equation.

$\begin{array}{c}{V}_1=\frac{\left|\begin{array}{cc}{I}_1& -j\left(1\times {10}^{-4}\right)\\ I_2& \left(1\times {10}^{-4}\right)+j\left(1\times {10}^{-4}\right)\end{array}\right|}{\left|\begin{array}{cc}{10}^{-5}+j\left(6\times {10}^{-4}\right)& -j\left(1\times {10}^{-4}\right)\\ 0.01-j\left(1\times {10}^{-4}\right)& \left(1\times {10}^{-4}\right)+j\left(1\times {10}^{-4}\right)\end{array}\right|}\\ =\frac{\left({10}^{-4}+j{10}^{-4}\right)I_1-\left(-j{10}^{-4}\right)I_2}{\left(-5.9\times {10}^{-8}+j6.1\times {10}^{-8}\right)-\left(-1\times {10}^{-8}-j{10}^{-6}\right)}\\ =\frac{\left({10}^{-4}+j{10}^{-4}\right)I_1+j{10}^{-4}{I}_2}{-\left(4.9\times {10}^{-8}\right)+j\left(1.061\times {10}^{-6}\right)}\\ =\frac{1.414\times {10}^{-4}\angle 45°I_1+{10}^{-4}\angle 90°I_2}{1.062\times {10}^{-6}\angle 92.64°}\end{array}$

$V_1=\left(89.71-j98.38\right)I_1+\left(94.06-j4.34\right)I_2$        (7)

The voltage expression $V_2$ by Crammer’s rule is given by,

$V_2=\frac{\left|\begin{array}{cc}{y}_{11}& I_1\\ y_{21}& I_2\end{array}\right|}{\left|\begin{array}{cc}{y}_{11}& y_{12}\\ y_{21}& y_{22}\end{array}\right|}$

Substitute ${10}^{-5}+j\left(6\times {10}^{-4}\right)$ for $y_{11}$, $-j\left(1\times {10}^{-4}\right)$ for $y_{12}$, $0.01-j\left(1\times {10}^{-4}\right)$ for $y_{21}$ and $\left(1\times {10}^{-4}\right)+j\left(1\times {10}^{-4}\right)$ for $y_{22}$ in above equation.

$\begin{array}{c}{V}_1=\frac{\left|\begin{array}{cc}{10}^{-5}+j\left(6\times {10}^{-4}\right)& I_1\\ 0.01-j\left(1\times {10}^{-4}\right)I_2& I_2\end{array}\right|}{\left|\begin{array}{cc}{10}^{-5}+j\left(6\times {10}^{-4}\right)& -j\left(1\times {10}^{-4}\right)\\ 0.01-j\left(1\times {10}^{-4}\right)& \left(1\times {10}^{-4}\right)+j\left(1\times {10}^{-4}\right)\end{array}\right|}\\ =\frac{{10}^{-5}+j\left(6\times {10}^{-4}\right)I_2-\left(0.01-j\left(1\times {10}^{-4}\right)\right)I_1}{\left(-5.9\times {10}^{-8}+j6.1\times {10}^{-8}\right)-\left(-1\times {10}^{-8}-j{10}^{-6}\right)}\\ =\frac{\left({10}^{-5}+j\left(6\times {10}^{-4}\right)\right)I_2-\left(0.01-j\left(1\times {10}^{-4}\right)\right)I_1}{-\left(4.9\times {10}^{-8}\right)+j\left(1.061\times {10}^{-6}\right)}\\ =\frac{6\times {10}^{-4}\angle 89.04°I_2+0.01\angle -0.57°I_1}{1.062\times {10}^{-6}\angle 92.64°}\end{array}$

$V_2=\left(527.27+j9401.42\right)I_1+\left(563.86-j35.47\right)I_2$        (8)

The standard equation for impedance parameter are,

$V_1=z_{11}{I}_1+z_{12}{I}_2$        (9)

$V_2=z_{21}{I}_1+z_{22}{I}_2$        (10)

Compare equation (7) with equation (9).

$z_{11}=\left(89.71-j98.38\right)\text{Ω}$

$z_{12}=\left(94.06-j4.34\right)\text{Ω}$

Compare equation (8) with equation (10).

$z_{21}=\left(527.27+j9401.42\right)\text{Ω}$

$z_{22}=\left(563.86-j35.47\right)\text{Ω}$

The $z$ matrix can be written as,

$\begin{array}{c}z=\left[\begin{array}{cc}\left(89.71-j98.38\right)\text{Ω}& \left(94.06-j4.34\right)\text{Ω}\\ \left(527.27+j9401.42\right)\text{Ω}& \left(563.86-j35.47\right)\text{Ω}\end{array}\right]\\ =\left[\begin{array}{cc}\left(89.71-j98.38\right)& \left(94.06-j4.34\right)\\ \left(527.27+j9401.42\right)& \left(563.86-j35.47\right)\end{array}\right]\text{Ω}\end{array}$

Conclusion:

Therefore, the required $z$ parameters is $z=\left[\begin{array}{cc}\left(89.71-j98.38\right)& \left(94.06-j4.34\right)\\ \left(527.27+j9401.42\right)& \left(563.86-j35.47\right)\end{array}\right]\text{Ω}$.